On Ramsey equilibrium: capital ownership pattern and inefficiency

Abstract

We provide a sufficient condition on the production function under which eventually the most patient household owns the entire capital stock in every Ramsey equilibrium, called the turnpike property. This generalizes the result in the literature which establishes the turnpike property using the capital income monotonicity condition. We then provide an example of a Ramsey equilibrium in which the most patient household reaches a no capital position infinitely often. This is a strong refutation of the turnpike property on Ramsey equilibria. We also show that the constructed Ramsey equilibrium is inefficient in terms of the aggregate consumption stream that it provides.

Appendix: Proofs

1.1 Proof of Theorem 4

Proof of Lemma 1

Since $\{K_{t-1}\}$ is not convergent, we have:

$$\begin{aligned} a\equiv \limsup _{t\rightarrow \infty } K_t >\liminf _{t\rightarrow \infty } K_t \equiv b. \end{aligned}$$

Denote $(a-b)$ by $\theta$. Since $\liminf _{t\rightarrow \infty } K_t = b$, we can pick $T_1 = \min \{t\in \mathbb N :K_{t} < b + \frac{\theta }{3}\}$. Since there is a subsequence of periods for which $\delta _{1} f^{\prime } (K_t)<1$, we have $K_t > K^{\delta _1}$ for that subsequence of periods. So, we can pick $T_2 = \min \{t\in \mathbb N : t > T_1\; \text {such that}\; K_{t} > K^{\delta _1}\}$. Since $a = \limsup _{t\rightarrow \infty } K_t$, we can pick $T_3 = \min \{t\in \mathbb N : t > T_2\; \text {such that}\; K_t > a - \frac{\theta }{3}\}$. And, since $\liminf _{t\rightarrow \infty } K_t = b$, we can pick $T_4 = \min \{t\in \mathbb N : t > T_3\; \text {such that}\; K_t < b + \frac{\theta }{3}\}$. Define $m = \max \{K_t: T_1\le t \le T_4\}$. Then,

$$\begin{aligned} \mathrm{(i)}\quad&m>K^{\delta _1};\nonumber \\ \mathrm{(ii)}\quad&m> a-(\theta /3) > b+(\theta /3). \end{aligned}$$

(20)

Now define $N = \max \{T_1 \le t \le T_4: K_t = m\}$. Then, using (20)(ii), we have $N < T_4$ and $N > T_1$. Further, by definition of $m$, we have $K_{N-1} \le K_N$ and by definition of $N$, we have $K_{N+1} < K_N$. Also, by (20)(i), $K_N> K^{\delta _1}$ and so $\delta _1 f^{\prime } (K_N) <1$. This completes the proof of the Lemma. $\square$

Proof of Lemma 2

Suppose the condition $K_{N-1} \le K_N$ and $K_{N+1}<K_N$ is satisfied for some $N \in \mathbb N$, but the condition $\delta _1 f^{\prime }(K_N)>1$ is violated. Then

$$\begin{aligned} \delta _1 f^{\prime }(K_N) \le 1. \end{aligned}$$

(21)

Let $\Gamma = \{h \in \{1,\ldots , H \}: x^h_N>0\}$. Denote the cardinality of $\Gamma$ by $\gamma$. Then for each $h\in \Gamma$, by (21),

$$\begin{aligned} \frac{u^{\prime }_h(c^h_{N+1})}{u^{\prime }_h(c^h_{N})}= \frac{1}{\delta _h f^{\prime }(K_N)}\ge 1. \end{aligned}$$

Thus, for each $h\in \Gamma$, we get $c^h_{N+1}\le c^h_{N}$, and this can be written as:

$$\begin{aligned} w_{N+1} + f^{\prime }(K_N) x^h_N - x^h_{N+1} \le w_{N} + f^{\prime }(K_{N-1}) x^h_{N-1}-x^h_N \;\text {for each} \; h\in \Gamma . \end{aligned}$$

(22)

Summing (22) over all $h\in \Gamma$, we obtain:

$$\begin{aligned}&\gamma w_{N+1} + f^{\prime }(K_N)\sum _{h\in \Gamma } x^h_N -\sum _{h\in \Gamma } x^h_{N+1} \le \gamma w_{N}\nonumber \\&\quad +f^{\prime }(K_{N-1}) \sum _{h\in \Gamma } x^h_{N-1}-\sum _{h\in \Gamma } x^h_{N}. \end{aligned}$$

(23)

By definition of $\Gamma$, we must have $\sum _{h\in \Gamma } x^h_{N}=K_N$, so that (23) can be written as:

$$\begin{aligned} \gamma w_{N+1} + f^{\prime }(K_N)K_N -\sum _{h\in \Gamma } x^h_{N+1} \le \gamma w_{N} + f^{\prime }(K_{N-1}) \sum _{h\in \Gamma } x^h_{N-1}-K_{N}. \end{aligned}$$

(24)

Clearly, $\sum _{h\in \Gamma } x^h_{N+1}\le K_{N+1}$ and $K_{N+1}<K_N$. So, (24) yields:

$$\begin{aligned} \gamma w_{N+1} + f^{\prime }(K_N)K_N&\le \gamma w_{N} + f^{\prime }(K_{N-1}) \sum _{h\in \Gamma } x^h_{N-1}-K_{N}+\sum _{h\in \Gamma } x^h_{N+1}\nonumber \\&\le \gamma w_{N} + f^{\prime }(K_{N-1}) \sum _{h\in \Gamma } x^h_{N-1}-K_{N}+K_{N+1}\nonumber \\&< \gamma w_{N} + f^{\prime }(K_{N-1})\sum _{h\in \Gamma } x^h_{N-1}. \end{aligned}$$

(25)

Then, since $\sum _{h\in \Gamma } x^h_{N-1}\le K_{N-1}$, (25) implies that:

$$\begin{aligned} \gamma w_{N+1} + f^{\prime }(K_N)K_N< \gamma w_{N} + f^{\prime }(K_{N-1})K_{N-1}. \end{aligned}$$

(26)

We can rewrite (26) as:

$$\begin{aligned} \gamma [w_{N+1}-w_N] < f^{\prime }(K_{N-1})K_{N-1}- f^{\prime }(K_N)K_N. \end{aligned}$$

Now, noting that $\gamma \ge 1$, and

$$\begin{aligned} w_{N+1}= \frac{f(K_N)-K_N f^{\prime }(K_N)}{H}\ge \frac{f(K_{N-1})-K_{N-1} f^{\prime }(K_{N-1})}{H} = w_N, \end{aligned}$$

(since $K_N \ge K_{N-1}$), we get,

$$\begin{aligned}{}[w_{N+1}-w_N]\le \gamma [w_{N+1}-w_N] < f^{\prime }(K_{N-1})K_{N-1}- f^{\prime }(K_N)K_N, \end{aligned}$$

(27)

and this can be rewritten as:

$$\begin{aligned} w_{N+1}+f^{\prime }(K_N)K_N< w_N+ f^{\prime }(K_{N-1})K_{N-1}. \end{aligned}$$

(28)

However, since $K_N\ge K_{N-1}$, (28) violates (15). This establishes the Lemma. $\square$

Proof of Theorem 4

We establish (i) as follows. Suppose, contrary to (i), $\{K_{t-1}\}$ is not convergent. If $\delta _{1} f^{\prime } (K_t)<1$ for a subsequence of periods, then by Lemma 1, there is $N\in \mathbb N$ such that:

$$\begin{aligned} \mathrm{(i)}\quad&K_{N-1}\le K_{N};\quad K_{N+1}<K_{N},\nonumber \\ \mathrm{(ii)}\quad&\delta _{1}f^{\prime }(K_{N})<1. \end{aligned}$$

(29)

However, since (15) holds, (29) (i) implies by Lemma 2 that $\delta _{1} f^{\prime } (K_N)>1$, which contradicts (29) (ii). Thus, there is $T\in \mathbb N$ such that

$$\begin{aligned} \delta _{1}f^{\prime }(K_{T})\ge 1\; \text {for all}\; t\ge T. \end{aligned}$$

(30)

Using the Ramsey–Euler inequality for household 1, we get,

$$\begin{aligned} \frac{u^{\prime }_1(c^1_{t+1})}{u^{\prime }_1(c^1_{t})}\le \frac{1}{\delta _1 f^{\prime }(K_t)}\; \text {for all}\; t\ge T. \end{aligned}$$

(31)

Thus, we must have,

$$\begin{aligned} c^1_{t+1}\ge c^1_{t}\; \text {for all}\; t\ge T. \end{aligned}$$

This implies that $\{c^1_t\}$ must converge to some $c^{\prime }>0$, and using this in (31), we have

$$\begin{aligned} \limsup _{t\rightarrow \infty }\; \delta _1 f^{\prime }(K_{t})\le 1. \end{aligned}$$

On the other hand, by (30), we have

$$\begin{aligned} \liminf _{t\rightarrow \infty }\; \delta _1 f^{\prime }(K_{t})\ge 1. \end{aligned}$$

Thus,

$$\begin{aligned} \liminf _{t\rightarrow \infty }\; \delta _1 f^{\prime }(K_{t}) = 1 = \limsup _{t\rightarrow \infty }\; \delta _1 f^{\prime }(K_{t}). \end{aligned}$$

But this means that $\{K_t\}$ is convergent, a contradiction. This establishes (i).

Using (i), we see that (ii) follows from Theorem 3. This establishes the Theorem. $\square$

1.2 Proof of Proposition 2

The following Lemma will be useful for the proof of Proposition 2.

Lemma 3

On any period three Ramsey equilibrium which is inefficient, the following capital stock ownership pattern must hold:

$$\begin{aligned} x^1_{T-1}>0; \quad x^1_T>0; \quad and \quad x^1_{T+1}=0. \end{aligned}$$

(32)

Proof

As a preliminary observation, we note that:

$$\begin{aligned} K_{T}>K^{g}. \end{aligned}$$

(33)

For if $K_{T}\le K^{g}$, then $K_{t}\le K^{g}$ for all $t\ge 0$, and so $f^{\prime }(K_{t})\ge 1$ for all $t\ge 0$. By the criterion of Cass (1972), the Ramsey equilibrium path must then be efficient, a contradiction.

Also, denoting the solution of $\delta _2f^{\prime }(k)=1$ by $\underline{K}$, and noting that the path is periodic, we have (by Proposition 2 of Becker and Foias (1987)):

$$\begin{aligned} K_{t}\ge \underline{K}\;\;\text {and }\; \delta _{2}f^{\prime }(K_{t})\le 1\;\;\text {for all }\; t\ge 0. \end{aligned}$$

(34)

If the conditions of the lemma do not hold, then consider the case where $x_{T}^{1}=0$, i.e., when the capital stock is at its maximum level, the patient agent does not hold any capital. In this case, $x_{T}^{2}=K_{T}>0$. Then,

$$\begin{aligned} c_{T}^{2}=w_{T}+(1+r_{T})x_{T-1}^{2}-x_{T}^{2} =w_{T}+(1+r_{T})x_{T-1}^{2}-K_{T}, \end{aligned}$$

(35)

and

$$\begin{aligned} c_{T+1}^{2}&= w_{T+1}+(1+r_{T+1})x_{T}^{2}-x_{T+1}^{2}=w_{T+1}+(1+r_{T+1})K_{T}-x_{T+1}^{2} \nonumber \\&\ge w_{T+1}+(1+r_{T+1})K_{T}-K_{T+1}. \end{aligned}$$

(36)

Further, by the choice of $K_T=M$, we have,

$$\begin{aligned} K_{T}\ge K_{T-1}, \end{aligned}$$

(37)

so that:

$$\begin{aligned} w_{T+1}&= [f(K_{T})-K_{T}f^{\prime }(K_{T})]/2 \nonumber \\&\ge [f(K_{T-1})-K_{T-1}f^{\prime }(K_{T-1})]/2 = w_{T} \end{aligned}$$

(38)

We have from (9):

$$\begin{aligned} \frac{u_{2}^{\prime }(c_{T+1}^{2})}{u_{2}^{\prime }(c_{T}^{2})}=\frac{1}{\delta _{2}f^{\prime }(K_{T})}. \end{aligned}$$

(39)

Thus, using (39) and (34) we have:

$$\begin{aligned} c_{T+1}^{2}\le c_{T}^{2}. \end{aligned}$$

(40)

So, by using (35), (36), (37), (38) and (40) we must have:

$$\begin{aligned} x_{T-1}^{2}>0. \end{aligned}$$

By periodicity and the recurrence theorem, we must also have:

$$\begin{aligned} x_{T+1}^{2}=0\;\;\text {and }x_{T+1}^{1}=K_{T+1}. \end{aligned}$$

Since $x_{T-1}^{2}>0$, we have from (9):

$$\begin{aligned} \frac{u_{2}^{\prime }(c_{T}^{2})}{u_{2}^{\prime }(c_{T-1}^{2})}=\frac{1}{\delta _{2}f^{\prime }(K_{T-1})}. \end{aligned}$$

(41)

Using (34), we get:

$$\begin{aligned} \delta _{2}f^{\prime }(K_{T-1})\le 1. \end{aligned}$$

(42)

Thus, using (41) and (42) we have:

$$\begin{aligned} c_{T}^{2}\le c_{T-1}^{2}. \end{aligned}$$

(43)

Combining (40) and (43), we obtain:

$$\begin{aligned} c_{T+1}^{2}\le c_{T-1}^{2}. \end{aligned}$$

(44)

Since $x_{T-2}^{2}=x_{T+1}^{2}=0$, we have:

$$\begin{aligned} c_{T-1}^{2}&= w_{T-1}+(1+r_{T-1})x_{T-2}^{2}-x_{T-1}^{2}=w_{T-1}-x_{T-1}^{2}\nonumber \\&= [f(K_{T-2})-K_{T-2}f^{\prime }(K_{T-2})]/2-x_{T-1}^{2}, \end{aligned}$$

and since $x_{T+1}^{2}=0$, we also have:

$$\begin{aligned} c_{T+1}^{2}&= w_{T+1}+(1+r_{T+1})x_{T}^{2}-x_{T+1}^{2}=w_{T+1}+(1+r_{T+1})K_{T} \nonumber \\&= [f(K_{T})-K_{T}f^{\prime }(K_{T})]/2+(1+r_{T+1})K_{T}. \end{aligned}$$

(45)

But, using the fact that:

$$\begin{aligned} K_{T}\ge K_{T+1}=K_{T-2}, \end{aligned}$$

(46)

and (45), (46), we must have:

$$\begin{aligned} c_{T+1}^{2}>c_{T-1}^{2}, \end{aligned}$$

which contradicts (44). Thus this possibility is ruled out.

The remaining possibility is $x^1_T>0$. In this case, we break up our analysis into four sub cases.

• (a) $x_{T-1}^{1}=x_{T+1}^{1}=0$;

• (b) $x_{T-1}^{1}>0$ and $x_{T+1}^{1}>0$;

• (c) $x_{T-1}^{1}=0$ and $x_{T+1}^{1}>0$; and

• (d) $x_{T-1}^{1}>0$ and $x_{T+1}^{1}=0$.

We proceed to show below that the cases (a), (b) and (c) cannot occur.

• (a) In this sub case,

  $$\begin{aligned} c_{T}^{1}=w_{T}+(1+r_{T})x_{T-1}^{1}-x_{T}^{1}=w_{T}-x_{T}^{1}<w_{T}, \end{aligned}$$

  (47)

  and

  $$\begin{aligned} c_{T+1}^{1}&= w_{T+1}+(1+r_{T+1})x_{T}^{1}-x_{T+1}^{1} \nonumber \\&= w_{T+1}+(1+r_{T+1})x_{T}^{1}>w_{T+1}. \end{aligned}$$

  (48)

  Further,

  $$\begin{aligned} K_{T}\ge K_{T-1}, \end{aligned}$$

  so that:

  $$\begin{aligned} w_{T+1}&= [f(K_{T})-K_{T}f^{\prime }(K_{T})]/2 \nonumber \\&\ge [f(K_{T-1})-K_{T-1}f^{\prime }(K_{T-1})]/2 = w_{T}. \end{aligned}$$

  (49)

  Thus, (47), (48) and (49) imply:

  $$\begin{aligned} c_{T+1}^{1}>c_{T}^{1}. \end{aligned}$$

  (50)

  Since $x_{T}^{1}>0$, we have from (9):

  $$\begin{aligned} \frac{u_{1}^{\prime }(c_{T+1}^{1})}{u_{1}^{\prime }(c_{T}^{1})}=\frac{1}{\delta _{1}f^{\prime }(K_{T})}. \end{aligned}$$

  (51)

  Using (33), we get:

  $$\begin{aligned} \delta _{1}f^{\prime }(K_{T})<1. \end{aligned}$$

  (52)

  Combining (51) and (52), we have:

  $$\begin{aligned} u_{1}^{\prime }(c_{T+1}^{1})>u_{1}^{\prime }(c_{T}^{1}), \end{aligned}$$

  which contradicts (50). Thus sub case (a) cannot arise.

• (b) In this sub case $x_{t}^{1}>0$ for all $t\ge 0$. Then, we have from (9):

  $$\begin{aligned} \frac{\delta _{1}u_{1}^{\prime }(c_{t+1}^{1})}{u_{1}^{\prime }(c_{t}^{1})} = \frac{1}{f^{\prime }(K_{t})}\ \ \text {for all }t\ge 1. \end{aligned}$$

  (53)

  Using the price sequence $\{p_{t}\}$ defined by (2), we obtain from (53), for all $\tau \ge 1$,

  $$\begin{aligned} \frac{p_{\tau +1}}{p_{0}}=\frac{\delta _{1}^{\tau }u_{1}^{\prime }(c_{\tau +1}^{1})}{u_{1}^{\prime }(c_{1}^{1})f^{\prime }(K_{0})}. \end{aligned}$$

  (54)

  Since $c_{\tau +1}^{1}\ge \eta$ for all $\tau \ge 1$, the right-hand side of (54) converges to zero as $\tau \rightarrow \infty$. Thus $p_{\tau +1}\rightarrow 0$ as $\tau \rightarrow \infty$, so that condition (4) is satisfied and the Ramsey equilibrium is efficient by Theorem 1. This contradiction shows that sub case (b) cannot arise.

• (c) We analyze this sub case by further subdividing into two parts (I) $x_{T}^{2}>0$; (II) $x_{T}^{2}=0$.

  • (I) Observe that agent 2 holds positive capital stock in periods $T-1$ and $T$. Then, by the recurrence theorem, we must have $x_{T+1}^{2}=0$. Using (9) for household 2, we get:

    $$\begin{aligned} \frac{u_{2}^{\prime }(c_{T+1}^{2})}{u_{2}^{\prime }(c_{T}^{2})} = \frac{1}{\delta _{2}f^{\prime }(K_{T})}\quad \text {and}\quad \frac{u_{2}^{\prime }(c_{T}^{2})}{u_{2}^{\prime }(c_{T-1}^{2})} = \frac{1}{\delta _{2}f^{\prime }(K_{T-1})}. \end{aligned}$$

    Using property (34), we obtain:

    $$\begin{aligned} c_{T+1}^{2}\le c_{T}^{2}\le c_{T-1}^{2}. \end{aligned}$$

    (55)

    However,

    $$\begin{aligned} c_{T+1}^{2}&= w_{T+1}+(1+r_{T+1})x_{T}^{2}-x_{T+1}^{2}=w_{T+1}+(1+r_{T+1})x_{T}^{2} \nonumber \\&> [f(K_{T})-K_{T}f^{\prime }(K_{T})]/2, \end{aligned}$$

    (56)

    and:

    $$\begin{aligned} c_{T-1}^{2}&= w_{T-1}+(1+r_{T-1})x_{T-2}^{2}-x_{T-1}^{2}=w_{T-1}-x_{T-1}^{2} \nonumber \\&< [f(K_{T-2})-K_{T-2}f^{\prime }(K_{T-2})]/2, \end{aligned}$$

    (57)

    since $x_{T-2}^{2}=x_{T+1}^{1}=0$. Then (56) and (57) contradict (55), since $K_{T}\ge K_{T-2}=K_{T+1}$.

  • (II) In this case, we use (9) for household 1 to get:

    $$\begin{aligned} \frac{u_{1}^{\prime }(c_{T+1}^{1})}{u_{1}^{\prime }(c_{T}^{1})}=\frac{1}{\delta _{1}f^{\prime }(K_{T})}. \end{aligned}$$

    (58)

    By (33), we have $\delta _{1}f^{\prime }(K_{T})<1$, and so (58) yields:

    $$\begin{aligned} c_{T+1}^{1}\le c_{T}^{1}. \end{aligned}$$

    (59)

    However,

    $$\begin{aligned} c_{T+1}^{1}&= w_{T+1}+(1+r_{T+1})x_{T}^{1}-x_{T+1}^{1} > w_{T+1}-x_{T+1}^{1} \nonumber \\&\ge [f(K_{T})-K_{T}f^{\prime }(K_{T})]/2-K_{T+1}, \end{aligned}$$

    (60)

    and:

    $$\begin{aligned} c_{T}^{1}&= w_{T}+(1+r_{T})x_{T-1}^{1}-x_{T}^{1} = w_{T}-x_{T}^{1} \nonumber \\&= [f(K_{T-1})-K_{T-1}f^{\prime }(K_{T-1})]/2-K_{T}, \end{aligned}$$

    (61)

    since $x_{T}^{2}=0$ and so $x_{T}^{1}=K_{T}$. Clearly (60) and (61) contradict (59) since $K_{T}\ge K_{T+1}$ and $K_{T}\ge K_{T-1}$. Thus, we can conclude that sub case (c) cannot arise.

The discussion of all the cases above leaves us with the sub case (d). Hence, for a period three Ramsey equilibrium to be inefficient, it is necessary that the conditions of the Lemma hold. $\square$

Proof of Proposition 2

Using Lemma 3, we know that (32) holds for any period three inefficient Ramsey equilibrium. We employ the two Euler equations (9) for household 1 (using $x_{T}^{1}>0$ and $x_{T-1}^{1}>0$ ) to get:

$$\begin{aligned} \frac{u_{1}^{\prime }(c_{T+1}^{1})}{u_{1}^{\prime }(c_{T}^{1})} = \frac{1}{\delta _{1}f^{\prime }(K_{T})} \quad \text {and} \quad \frac{u_{1}^{\prime }(c_{T}^{1})}{u_{1}^{\prime }(c_{T-1}^{1})} = \frac{1}{\delta _{1}f^{\prime }(K_{T-1})}. \end{aligned}$$

This yields:

$$\begin{aligned} \frac{u_{1}^{\prime }(c_{T+1}^{1})}{u_{1}^{\prime }(c_{T-1}^{1})} = \frac{u_{1}^{\prime }(c_{T+1}^{1})}{u_{1}^{\prime }(c_{T}^{1})}\frac{u_{1}^{\prime }(c_{T}^{1})}{u_{1}^{\prime }(c_{T-1}^{1})} = \frac{1}{(\delta _{1})^{2}f^{\prime }(K_{T})f^{\prime }(K_{T-1})}. \end{aligned}$$

(62)

Note that since $x_{T+1}^{1}\equiv x_{T-2}^{1}=0$, we have:

$$\begin{aligned} c_{T+1}^{1}&= w_{T+1}+(1+r_{T+1})x_{T}^{1}-x_{T+1}^{1} = w_{T+1}+(1+r_{T+1})x_{T}^{1} \nonumber \\&> [f(K_{T})-K_{T}f^{\prime }(K_{T})]/2, \end{aligned}$$

(63)

and:

$$\begin{aligned} c_{T-1}^{1}&= w_{T-1}+(1+r_{T-1})x_{T-2}^{1}-x_{T-1}^{1} = w_{T-1}-x_{T-1}^{1} \nonumber \\&< [f(K_{T-2})-K_{T-2}f^{\prime }(K_{T-2})]/2. \end{aligned}$$

(64)

Using (63) and (64), we infer:

$$\begin{aligned} c_{T+1}^{1}>c_{T-1}^{1}, \end{aligned}$$

(65)

since $K_{T}\ge K_{T+1}\equiv K_{T-2}$. Using (65) and (62), we obtain:

$$\begin{aligned} (\delta _{1})^{2}f^{\prime }(K_{T})f^{\prime }(K_{T-1})>1. \end{aligned}$$

(66)

Using (33) and (66), we get:

$$\begin{aligned} f^{\prime }(K_{T-1})>\frac{1}{(\delta _{1})^{2}}>1. \end{aligned}$$

Thus, we must have:

$$\begin{aligned} K_{T-1}< K^{g}. \end{aligned}$$

(67)

Using (66) again, we have:

$$\begin{aligned} (\delta _{1})^{2}f^{\prime }(K_{T+1})f^{\prime }(K_{T})f^{\prime }(K_{T-1})>f^{\prime }(K_{T+1}), \end{aligned}$$

(68)

and by the Cass criterion (6) of inefficiency, we must have:

$$\begin{aligned} 1>f^{\prime }(K_{T+1})f^{\prime }(K_{T})f^{\prime }(K_{T-1}). \end{aligned}$$

(69)

Using (69) in (68) yields:

$$\begin{aligned} 1>(\delta _{1})^{2}>f^{\prime }(K_{T+1}), \end{aligned}$$

(70)

and so:

$$\begin{aligned} K_{T+1}> K^{g}. \end{aligned}$$

(71)

We now use (67) and (71) to examine the behavior of household 2. In particular, we claim that:

$$\begin{aligned} x_{T}^{2}=0. \end{aligned}$$

(72)

For, if (72) is violated, then we have $x_{T}^{2}>0$ and also $x_{T+1}^{2}=K_{T+1}>0$ (since $x_{T+1}^{1}=0$ by Lemma 3), so that by the recurrence theorem we have:

$$\begin{aligned} x_{T-1}^{2}=0, \end{aligned}$$

(73)

and the two Euler equations (9) for household 2 yield:

$$\begin{aligned} \frac{u_{2}^{\prime }(c_{T+1}^{2})}{u_{2}^{\prime }(c_{T}^{2})}=\frac{1}{\delta _{2}f^{\prime }(K_{T})}\ge 1\quad \text {and}\quad \frac{u_{2}^{\prime }(c_{T+2}^{2})}{u_{2}^{\prime }(c_{T+1}^{2})}=\frac{1}{\delta _{2}f^{\prime }(K_{T+1})}\ge 1, \end{aligned}$$

using observation (34). Thus, we obtain:

$$\begin{aligned} c_{T-1}^{2}\equiv c_{T+2}^{2}\le c_{T+1}^{2}\le c_{T}^{2}. \end{aligned}$$

(74)

However, we have:

$$\begin{aligned} c_{T}^{2}&= w_{T}+(1+r_{T})x_{T-1}^{2}-x_{T}^{2}=w_{T}-x_{T}^{2} \nonumber \\&< [f(K_{T-1})-K_{T-1}f^{\prime }(K_{T-1})]/2, \end{aligned}$$

and:

$$\begin{aligned} c_{T-1}^{2}&= w_{T-1}+(1+r_{T-1})x_{T-2}^{2}-x_{T-1}^{2}=w_{T-1}+(1+r_{T-1})x_{T-2}^{2} \nonumber \\&> [f(K_{T-2})-K_{T-2}f^{\prime }(K_{T-2})]/2, \end{aligned}$$

using (73). Since $K_{T-2}\equiv K_{T+1}>K_{T-1}$ by (67) and (71), we must have:

$$\begin{aligned} c_{T-1}^{2}>[f(K_{T-2})-K_{T-2}f^{\prime }(K_{T-2})]/2>[f(K_{T-1})-K_{T-1}f^{\prime }(K_{T-1})]/2>c_{T}^{2}, \end{aligned}$$

and this contradicts (74). This establishes our claim (72) and therefore part (b) of the Proposition.

Part (a) of the Proposition will be established, given (67) and (71) if we can prove that:

$$\begin{aligned} K_{T}>K_{T+1}. \end{aligned}$$

(75)

If (75) were violated, then $K_{T}=K_{T+1}$, and further since $x_{T+1}^{2}=K_{T+1}>0$, (9) for household 2 yields:

$$\begin{aligned} \frac{u_{2}^{\prime }(c_{T+2}^{2})}{u_{2}^{\prime }(c_{T+1}^{2})}=\frac{1}{\delta _{2}f^{\prime }(K_{T+1})}\ge 1, \end{aligned}$$

(76)

using observation (34). Thus, we must have:

$$\begin{aligned} c_{T-1}^{2}\equiv c_{T+2}^{2}\le c_{T+1}^{2}. \end{aligned}$$

(77)

However, we have:

$$\begin{aligned} c_{T+1}^{2}&= w_{T+1}+(1+r_{T+1})x_{T}^{2}-x_{T+1}^{2}=w_{T+1}-x_{T+1}^{2} \nonumber \\&= [f(K_{T})-K_{T}f^{\prime }(K_{T})]/2-K_{T+1}, \end{aligned}$$

and:

$$\begin{aligned} c_{T+2}^{2}&= w_{T+2}+(1+r_{T+2})x_{T+1}^{2}-x_{T+2}^{2}>w_{T+2}-x_{T+2}^{2} \nonumber \\&\ge [f(K_{T+1})-K_{T+1}f^{\prime }(K_{T+1})]/2-K_{T-1}. \end{aligned}$$

So, using $K_{T}=K_{T+1}$ and $K_{T+1}>K_{T-1}$, we get $c_{T+2}^{2}>c_{T+1}^{2}$, contradicting (77). This establishes (75) and therefore part (a) of the Proposition.

We note that, having established (a), (76) can be strengthened to read:

$$\begin{aligned} \frac{u_{2}^{\prime }(c_{T+2}^{2})}{u_{2}^{\prime }(c_{T+1}^{2})}=\frac{1}{\delta _{2}f^{\prime }(K_{T+1})}> 1, \end{aligned}$$

so that we have a strengthened version of (77):

$$\begin{aligned} c_{T-1}^{2}\equiv c_{T+2}^{2}<c_{T+1}^{2}. \end{aligned}$$

(78)

To establish part (c) of the Proposition, we use (70) and (75) to infer that:

$$\begin{aligned} 1>(\delta _{1})^{2}>f^{\prime }(K_{T+1})>f^{\prime }(K_{T}). \end{aligned}$$

(79)

Then, we use (66) and (79) to obtain:

$$\begin{aligned} (\delta _{1})^{4}f^{\prime }(K_{T-1})>(\delta _{1})^{2}f^{\prime }(K_{T})f^{\prime }(K_{T-1})>1. \end{aligned}$$

(80)

By observation (34) we also have:

$$\begin{aligned} 1\ge \delta _{2}f^{\prime }(K_{T-1}). \end{aligned}$$

(81)

Combining (80) and (81), we obtain:

$$\begin{aligned} \delta _{2}<(\delta _{1})^{4}. \end{aligned}$$

which establishes part (c) of the Proposition.

To prove part (d), we note that since $x_{T}^{1}=K_{T}>K^{g}$, we can employ (9) for household 1 to get:

$$\begin{aligned} \frac{u_{1}^{\prime }(c_{T+1}^{1})}{u_{1}^{\prime }(c_{T}^{1})}=\frac{1}{\delta _{1}f^{\prime }(K_{T})}>1. \end{aligned}$$

This yields:

$$\begin{aligned} c_{T+1}^{1}<c_{T}^{1}. \end{aligned}$$

(82)

Thus, combining (65) and (82), we obtain:

$$\begin{aligned} c_{T-1}^{1}<c_{T+1}^{1}<c_{T}^{1}, \end{aligned}$$

which establishes (d).

To establish part (e), we make an observation about the aggregate consumption in periods $T$ and $T+1$,

$$\begin{aligned} c_{T}^{1}+c_{T}^{2}&=C_T=f(K_{T-1})-K_{T} < f(K_{T})-K_{T} \nonumber \\&<f(K_{T})-K_{T+1} = C_{T+1}=c_{T+1}^{1}+c_{T+1}^{2}. \end{aligned}$$

(83)

Using (82) and (83), we get:

$$\begin{aligned} c_{T+1}^{2}>c_{T}^{2}. \end{aligned}$$

(84)

Combining (78) and (84), we obtain:

$$\begin{aligned} c_{T+1}^{2}>\max \{c_{T}^{2},c_{T-1}^{2}\}. \end{aligned}$$

which establishes (e).

Finally, to establish (f), we proceed as follows. Note that since $x_{T+1}^{1}=0$, we have:

$$\begin{aligned} c_{T+1}^{1}&= w_{T+1}+(1+r_{T+1})x_{T}^{1}-x_{T+1}^{1} \nonumber \\&= w_{T+1}+(1+r_{T+1})K_{T}, \end{aligned}$$

(85)

and:

$$\begin{aligned} c_{T}^{1}&= w_{T}+(1+r_{T})x_{T-1}^{1}-x_{T}^{1} \nonumber \\&\le w_{T}+(1+r_{T})K_{T-1}-K_{T}. \end{aligned}$$

(86)

Using (82), (85) and (86), we infer:

$$\begin{aligned} (1+r_{T})K_{T-1}>K_{T}+(1+r_{T+1})K_{T}+(w_{T+1}-w_{T}). \end{aligned}$$

(87)

Since $x_{T}^{2}=0$ while $x_{T+1}^{2}=K_{T+1}$, we have:

$$\begin{aligned} c_{T+1}^{2}&= w_{T+1}+(1+r_{T+1})x_{T}^{2}-x_{T+1}^{2} = w_{T+1}-K_{T+1}, \end{aligned}$$

(88)

and:

$$\begin{aligned} c_{T}^{2}&= w_{T}+(1+r_{T})x_{T-1}^{2}-x_{T}^{2} \nonumber \\&= w_{T}+(1+r_{T})x_{T-1}^{2} \ge w_{T}. \end{aligned}$$

(89)

Using (84), (88) and (89), we obtain:

$$\begin{aligned} K_{T+1}<w_{T+1}-w_{T}. \end{aligned}$$

(90)

Combining (87) and (90), we get:

$$\begin{aligned} (1+r_{T})K_{T-1}>K_{T}+(1+r_{T+1})K_{T}+K_{T+1}. \end{aligned}$$

(91)

Since $K_{T-1}<K_{T+1}$, while $K_{T}>K_{T+1}$, the inequality in (91) yields:

$$\begin{aligned} (1+r_{T})K_{T+1}>K_{T+1}+(1+r_{T+1})K_{T+1}+K_{T+1}. \end{aligned}$$

(92)

Thus, we obtain:

$$\begin{aligned} f^{\prime }(K_{T-1}) =(1+r_{T})>1+(1+r_{T+1})+1 = 2+f^{\prime }(K_{T}). \end{aligned}$$

(93)

Using (66) in (93), we obtain finally:

$$\begin{aligned} f^{\prime }(K_{T-1})>\left[ 2+\frac{1}{f^{\prime }(K_{T-1})}\right] . \end{aligned}$$

(94)

Define $v(z)=$ $z^{2}-2z-1$ for all $z\in \mathbb R$. Then, $v(z)>0$ and $z>0$ implies $z>\sqrt{2}+1$. Thus, (94) implies that:

$$\begin{aligned} f^{\prime }(K_{T-1})>\sqrt{2}+1. \end{aligned}$$

Since $\delta _{2}f^{\prime }(K_{T-1})\le 1$, we must have:

$$\begin{aligned} \delta _{2}<\frac{1}{\sqrt{2}+1}=\sqrt{2}-1\approx 0.414. \end{aligned}$$

$\square$

1.3 Example of production function satisfying MIM

Consider the class of CES production functions given by

$$\begin{aligned} G(K,L) = A[\alpha K^{-\rho } +(1-\alpha )L^{-\rho }]^{-\frac{1}{\rho }}\;\text {for all}\; (K,L)\gg 0, \end{aligned}$$

(95)

where $A > 0$, $\alpha \in (0,1)$ and $\rho >0$. In this case, the (constant) elasticity of substitution is given by:

$$\begin{aligned} \sigma = \frac{1}{1+\rho }<1. \end{aligned}$$

(96)

The Cobb–Douglas production function is obtained as a limiting case by letting $\rho \rightarrow 0$, and so $\sigma \rightarrow 1$.

Since each household supplies one unit of labor, we can define the function $g(K)$ as:

$$\begin{aligned} g(K)= G(K,H) = A[\alpha K^{-\rho } +(1-\alpha )H^{-\rho }]^{-\frac{1}{\rho }}\;\text {for all}\; K>0, \end{aligned}$$

(97)

so that we have, for all $K>0$,

$$\begin{aligned} g(K)^{-\rho }= A^{-\rho }\alpha K^{-\rho } +A^{-\rho }(1-\alpha ) H^{-\rho }= a K^{-\rho }+b, \end{aligned}$$

(98)

where $a\equiv A^{-\rho }\alpha$ and $b\equiv A^{-\rho } (1-\alpha ) H^{-\rho }$.

Note that as $K\rightarrow 0$, we have $g(K)\rightarrow 0$. So, we can define $g(0)=0$ and preserve continuity of g on $\mathbb R _+$. The difficulty is that $g^{\prime }(K)$ is bounded as $K\rightarrow 0$. While one can ensure that the marginal product is very high for $K$ near 0 (by taking $A$ large), one does not know whether the existence of a Ramsey equilibrium can be ensured from every positive initial capital stock, and initial ownership pattern of the capital stock.

So, we consider (instead of CES) the approximate CES production function, given by:

$$\begin{aligned} f(K)= \left\{ \begin{array}{ll} \theta g(K)+ (1-\theta ) s(K)&{} \; \text {for all}\; K>0\\ 0&{} \; \text {for}\; K=0, \end{array} \right. \end{aligned}$$

where $\theta \in (0,1)$, and $s(K)$ is a (reduced-form) function,

$$\begin{aligned} s(K)= B K^{\beta } \; \text {for all}\; K>0, \end{aligned}$$

(99)

obtained from the Cobb–Douglas production function, with $B>0$ and $\beta \in (0,1)$. Then, $f$ satisfies all the maintained assumptions of our basic framework. For $\theta \approx 1$, $f$ is an approximate CES function. Note that we have:

$$\begin{aligned} \mathrm{(i)} \quad&f^{\prime }(K) = \theta g^{\prime }(K) +(1-\theta ) s^{\prime }(K)\; \text {for all}\; K>0\nonumber \\ \mathrm{(ii)} \quad&f^{\prime \prime }(K) = \theta g^{\prime \prime }(K) +(1-\theta ) s^{\prime \prime }(K)\; \text {for all}\; K>0. \end{aligned}$$

Thus, we have:

$$\begin{aligned} \frac{K[-f^{\prime \prime }(K)]}{f^{\prime }(K)}&= \frac{K\theta [-g^{\prime \prime }(K)]+K(1-\theta ) [-s^{\prime \prime }(K)]}{\theta g^{\prime }(K) +(1-\theta ) s^{\prime }(K)}\nonumber \\&\le \frac{K\theta [-g^{\prime \prime }(K)]}{\theta g^{\prime }(K)} + \frac{K(1-\theta ) [-s^{\prime \prime }(K)]}{(1-\theta ) s^{\prime }(K)}\nonumber \\&= \frac{K[-g^{\prime \prime }(K)]}{g^{\prime }(K)} + \frac{K[-s^{\prime \prime }(K)]}{s^{\prime }(K)}. \end{aligned}$$

Differentiating (98) with respect to $K$, and simplifying, we obtain:

$$\begin{aligned} g^{\prime }(K) = a \left[ \frac{g(K)}{K}\right] ^{\rho +1}\; \text {for all}\; K>0. \end{aligned}$$

(100)

Differentiating (100) with respect to $K$, we obtain,

$$\begin{aligned} g^{\prime \prime }(K) = a(\rho +1) \left[ \frac{g(K)}{K}\right] ^{\rho } \frac{K g^{\prime }(K) - g(K)}{K^2}. \end{aligned}$$

(101)

Using (100) and (101), we can simplify to obtain,

$$\begin{aligned} e_g(K) \equiv \left[ \frac{-K g^{\prime \prime }(K)}{g^{\prime }(K)}\right] = (\rho +1) \left[ 1-\frac{K g^{\prime }(K)}{g(K)}\right] < \rho +1 \end{aligned}$$

since $\left[ 1- \frac{K g^{\prime }(K)}{g(K)}\right] \in (0,1)$. Also,

$$\begin{aligned} e_s(K) \equiv \left[ \frac{-K s^{\prime \prime }(K)}{s^{\prime }(K)}\right] = 1-\beta . \end{aligned}$$

Thus, (MIM) holds if:

$$\begin{aligned} \left[ (\rho +1) + (1-\beta )\right] \left[ 1-\frac{1}{H}\right] <1, \end{aligned}$$

which can be rewritten as:

$$\begin{aligned} 0 < \rho +1 < \left[ \frac{H}{H-1}\right] - (1-\beta ). \end{aligned}$$

Since the constant elasticity of substitution of g is $\sigma _g = [1/(1+\rho )]$, (MIM) holds if:

$$\begin{aligned} \frac{1}{\sigma _g} < \frac{H}{H-1} - (1-\beta ). \end{aligned}$$

(102)

By choosing $\beta$ sufficiently close to 1 in the function $s(K)$, (102) can be made to hold if:

$$\begin{aligned} \frac{1}{\sigma _g} < \frac{H}{H-1}\quad \text {or}\quad \sigma _g> 1-\frac{1}{H}. \end{aligned}$$

This agrees with the formula derived earlier in Sect. 4.2.

1.4 Proof of inefficiency

Without loss of generality, consider the case where $k_0=\frac{174}{175}$. The equilibrium capital stock and aggregate consumption sequences in the example in Sect. 5.2 are:

$$\begin{aligned} \{K_t, C_{t+1}\} ={\left\{ \begin{array}{ll} &{}\left\{ \frac{174}{175}, c+132\right\} ,\;\text {for}\; t=0, 3, 6, \ldots \\ &{}\left\{ 42, c+177+\frac{13}{36}-\frac{1}{39}\right\} ,\; \text {for}\; t=1, 4, 7, \ldots \\ &{}\left\{ \frac{40}{39}, c+174+\frac{1}{13\cdot 36}+\frac{1}{175}\right\} ,\; \text {for}\; t=2, 5, 8, \ldots \end{array}\right. } \end{aligned}$$

(103)

We show that the capital stock sequence $\{K_t^\prime \}$ (described below) is feasible from $k_0$ and dominates $\{K_t\}$ as it provides higher aggregate consumption in period $1$ and same aggregate consumption in all the remaining periods. We take $\varepsilon =\frac{1}{175}$. The dominating capital stock sequence is:

$$\begin{aligned} K^\prime _t= {\left\{ \begin{array}{ll} &{}\frac{174}{175},\;\text {for}\; t=0,\\ &{}42-\left( \frac{175}{432}\right) ^n\cdot \left( \varepsilon \right) >41,\; \text {for}\; t=3n+1,\\ &{}\frac{40}{39}-\left( \frac{175}{432}\right) ^n\cdot \left( \frac{\varepsilon }{36}\right) >1,\; \text {for}\; t=3n+2,\\ &{}\frac{174}{175}-\left( \frac{175}{432}\right) ^n\cdot \left( \frac{\varepsilon }{432}\right) >\mu =\frac{1}{350^2},\; \text {for}\; t=3n+3, \end{array}\right. } \end{aligned}$$

(104)

where $n=0,1,\ldots$. Using the capital stock sequence given in (104), we can compute the aggregate output sequence as:

$$\begin{aligned} y^{\prime }_{t+1}=f(K^\prime _{t})= {\left\{ \begin{array}{ll} &{}c+174,\;\text {for}\; t=0,\\ &{}c+ 178 +\frac{13}{36}- \left( \frac{175}{432}\right) ^n\cdot \left( \frac{\varepsilon }{36}\right) ,\; \text {for}\; t=3n+1,\\ &{}c+ 175+\frac{1}{13\cdot 36}-\left( \frac{175}{432}\right) ^n\cdot \left( \frac{\varepsilon }{432}\right) ,\; \text {for}\; t=3n+2,\\ &{}c+174-\left( \frac{175}{432}\right) ^{n+1}\cdot \left( \varepsilon \right) ,\; \text {for}\; t=3n+3, \end{array}\right. } \end{aligned}$$

(105)

where $n=0,1,\ldots$. The corresponding aggregate consumption sequence is obtained from (104) and (105) as:

$$\begin{aligned}&C^{\prime }_{t+1} = y^{\prime }_{t+1}- K^{\prime }_{t+1}\nonumber \\&\quad =f(K^\prime _{t})-K^{\prime }_{t+1}= {\left\{ \begin{array}{ll} &{}c+132+\varepsilon ,\;\text {for}\; t=0,\\ &{}c+ 177 +\frac{13}{36}- \frac{1}{39},\; \text {for}\; t=3n+1,\\ &{}c+ 174+\frac{1}{13\cdot 36}+\frac{1}{175},\; \text {for}\; t=3n+2,\\ &{}c+132,\; \text {for}\; t=3n+3, \end{array}\right. } \end{aligned}$$

(106)

where $n=0,1,\ldots$. The sequences described above have been constructed as under:

• (a) Take $C_1^{\prime }=C_1+\varepsilon$ and $C_t^{\prime }=C_t$ $\forall t>1$. Then $K_1^{\prime }=K_1-\varepsilon =42-\frac{1}{175}>41$ and therefore $f(K_1^{\prime })=f(K_1)-\frac{\varepsilon }{36}=C_2+K_2- \left( \frac{\varepsilon }{36}\right)$.

• (b) $K_2^{\prime }=K_2-\left( \frac{\varepsilon }{36}\right) =\frac{40}{39}- \frac{1}{175}\cdot \frac{1}{36}=1+\frac{1}{39}- \frac{1}{175}\cdot \frac{1}{36}>1$. Also $f(K_2^{\prime })=f(K_2)-\left( \frac{\varepsilon }{36}\right) \cdot \frac{3}{36}=f(K_2)-\left( \frac{\varepsilon }{432}\right) = C_3+K_3-\left( \frac{\varepsilon }{432}\right)$.

• (c) $K_3^{\prime }=K_3-\left( \frac{\varepsilon }{432}\right) =k_0-\left( \frac{\varepsilon }{432}\right) =\frac{174}{175}- \frac{1}{175}\cdot \frac{1}{432}>\mu$. Further, $f(K_3^{\prime })=f(K_3)-\left( \frac{\varepsilon }{432}\right) \cdot 175=f(k_0)-\left( \frac{175}{432}\right) \cdot \varepsilon =C_4+K_4-\left( \frac{175}{432}\right) \cdot \varepsilon =C_1+K_1-\left( \frac{175}{432}\right) \cdot \varepsilon >f(k_0)-\varepsilon$.

• (d) $K_4^{\prime }=K_1-\left( \frac{175}{432}\right) \cdot \varepsilon >K_1-\varepsilon =K_1^\prime >41$. Then $f(K_4^{\prime })=f(K_1)-\frac{\left( \frac{175}{432}\right) \cdot \varepsilon }{36}=C_5+K_5-\left( \frac{175}{432}\right) \cdot \left( \frac{\varepsilon }{36}\right)$.

• (e) $K_5^{\prime }=K_5-\left( \frac{175}{432}\right) \cdot \frac{\varepsilon }{36}>K_2-\frac{\varepsilon }{36}=K_2^\prime >1$. Then $f(K_5^{\prime })=f(K_5)-\left( \frac{175}{432}\right) \cdot \left( \frac{\varepsilon }{36}\right) \cdot \frac{3}{36}=C_6+K_6-\left( \frac{175}{432}\right) \cdot \left( \frac{\varepsilon }{432}\right)$.

• (f) $K_6^{\prime }=K_6-\left( \frac{175}{432}\right) \cdot \left( \frac{\varepsilon }{432}\right) =K_3-\left( \frac{175}{432}\right) \cdot \left( \frac{\varepsilon }{432}\right) > K_3-\frac{\varepsilon }{432}=K_3^\prime$.

Following the same logic, we can construct the entire sequence. Comparing (103) and (106), we get $C_1^{\prime }=C_1+\varepsilon$ and $C_t^\prime =C_t$ for all $t>1$ and therefore $\{K_t\}$ is not efficient. In case, $k_0\ne \frac{174}{175}$, we initiate the construction of dominating sequence from $t=2$ for $k_0=42$ and $t=1$ for $k_0=\frac{40}{39}$.